Problem: Takahashi loves sorting.

He has a permutation (p_1,p_2,...,p_N) of the integers from 1 through N. Now, he will repeat the following operation until the permutation becomes (1,2,...,N):

* First, we will define high and low elements in the permutation, as follows. The i-th element in the permutation is high if the maximum element between the 1-st and i-th elements, inclusive, is the i-th element itself, and otherwise the i-th element is low.
* Then, let a_1,a_2,...,a_k be the values of the high elements, and b_1,b_2,...,b_{N-k} be the values of the low elements in the current permutation, in the order they appear in it.
* Lastly, rearrange the permutation into (b_1,b_2,...,b_{N-k},a_1,a_2,...,a_k).



How many operations are necessary until the permutation is sorted?

Constraints

* 1 ≤ N ≤ 2×10^5
* (p_1,p_2,...,p_N) is a permutation of the integers from 1 through N.

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 ... p_N


Output

Print the number of operations that are necessary until the permutation is sorted.

Examples

Input

5
3 5 1 2 4


Output

3


Input

5
5 4 3 2 1


Output

4


Input

10
2 10 5 7 3 6 4 9 8 1


Output

6

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

const int N = 202020;

inline char get(void) {
	static char buf[100000], *S = buf, *T = buf;
	if (S == T) {
		T = (S = buf) + fread(buf, 1, 100000, stdin);
		if (S == T) return EOF;
	}
	return *S++;
}
template<typename T>
inline void read(T &x) {
	static char c; x = 0; int sgn = 0;
	for (c = get(); c < '0' || c > '9'; c = get()) if (c == '-') sgn = 1;
	for (; c >= '0' && c <= '9'; c = get()) x = x * 10 + c - '0';
	if (sgn) x = -x;
}

int n;
int p[N], q[N];
int f[N], T[N];

inline bool Check(int a, int b, int c) {
	return (a < b && b < c) || (b < c && c < a) || (c < a && a < b);
}

int main(void) {
	read(n);
	for (int i = 1; i <= n; i++) {
		read(p[i]); q[p[i]] = i;
	}
	T[n] = 0;
	for (int i = n - 1; i; i--) {
		if (T[i + 1] == 0) {
			if (q[i] > q[i + 1]) {
				T[i] = 1; f[i] = i + 1;
			}
		} else {
			if (Check(q[f[i + 1]], q[i], q[i + 1])) {
				T[i] = T[i + 1]; f[i] = f[i + 1];
			} else {
				T[i] = T[i + 1] + 1; f[i] = i + 1;
			}
		}
	}
	cout << T[1] << endl;
}